以下是我从“cuda by example”一书中编辑的代码，用于测试 CUDA 并发内核执行。
static void HandleError( cudaError_t err,
const char *file,
int line ) {
if (err != cudaSuccess) {
printf( "%s in %s at line %d\n", cudaGetErrorString( err ),
file, line );
exit( EXIT_FAILURE );
}
}
#define HANDLE_ERROR( err ) (HandleError( err, __FILE__, __LINE__ ))
#define N (1024*1024*10)
#define FULL_DATA_SIZE (N*2)
__global__ void kernel( int *a, int *b, int *c ) {
int idx = threadIdx.x + blockIdx.x * blockDim.x;
if (idx < N) {
int idx1 = (idx + 1) % 256;
int idx2 = (idx + 2) % 256;
float as = (a[idx] + a[idx1] + a[idx2]) / 3.0f;
float bs = (b[idx] + b[idx1] + b[idx2]) / 3.0f;
c[idx] = (as + bs) / 2;
}
}
int main( void ) {
cudaDeviceProp prop;
int whichDevice;
HANDLE_ERROR( cudaGetDevice( &whichDevice ) );
HANDLE_ERROR( cudaGetDeviceProperties( &prop, whichDevice ) );
if (!prop.deviceOverlap) {
printf( "Device will not handle overlaps, so no speed up from streams\n" );
return 0;
}
cudaEvent_t start, stop;
float elapsedTime;
cudaStream_t stream0, stream1;
int *host_a, *host_b, *host_c;
int *dev_a0, *dev_b0, *dev_c0;
int *dev_a1, *dev_b1, *dev_c1;
// start the timers
HANDLE_ERROR( cudaEventCreate( &start ) );
HANDLE_ERROR( cudaEventCreate( &stop ) );
// initialize the streams
HANDLE_ERROR( cudaStreamCreate( &stream0 ) );
HANDLE_ERROR( cudaStreamCreate( &stream1 ) );
// allocate the memory on the GPU
HANDLE_ERROR( cudaMalloc( (void**)&dev_a0,
N * sizeof(int) ) );
HANDLE_ERROR( cudaMalloc( (void**)&dev_b0,
N * sizeof(int) ) );
HANDLE_ERROR( cudaMalloc( (void**)&dev_c0,
N * sizeof(int) ) );
HANDLE_ERROR( cudaMalloc( (void**)&dev_a1,
N * sizeof(int) ) );
HANDLE_ERROR( cudaMalloc( (void**)&dev_b1,
N * sizeof(int) ) );
HANDLE_ERROR( cudaMalloc( (void**)&dev_c1,
N * sizeof(int) ) );
// allocate host locked memory, used to stream
HANDLE_ERROR( cudaHostAlloc( (void**)&host_a,
FULL_DATA_SIZE * sizeof(int),
cudaHostAllocDefault ) );
HANDLE_ERROR( cudaHostAlloc( (void**)&host_b,
FULL_DATA_SIZE * sizeof(int),
cudaHostAllocDefault ) );
HANDLE_ERROR( cudaHostAlloc( (void**)&host_c,
FULL_DATA_SIZE * sizeof(int),
cudaHostAllocDefault ) );
for (int i=0; i<FULL_DATA_SIZE; i++) {
host_a[i] = rand();
host_b[i] = rand();
}
HANDLE_ERROR( cudaEventRecord( start, 0 ) );
// now loop over full data, in bite-sized chunks
for (int i=0; i<FULL_DATA_SIZE; i+= N*2) {
// enqueue kernels in stream0 and stream1
kernel<<<N/256,256,0,stream0>>>( dev_a0, dev_b0, dev_c0 );
kernel<<<N/256,256,0,stream1>>>( dev_a1, dev_b1, dev_c1 );
}
HANDLE_ERROR( cudaStreamSynchronize( stream0 ) );
HANDLE_ERROR( cudaStreamSynchronize( stream1 ) );
HANDLE_ERROR( cudaEventRecord( stop, 0 ) );
HANDLE_ERROR( cudaEventSynchronize( stop ) );
HANDLE_ERROR( cudaEventElapsedTime( &elapsedTime,
start, stop ) );
printf( "Time taken: %3.1f ms\n", elapsedTime );
// cleanup the streams and memory
HANDLE_ERROR( cudaFreeHost( host_a ) );
HANDLE_ERROR( cudaFreeHost( host_b ) );
HANDLE_ERROR( cudaFreeHost( host_c ) );
HANDLE_ERROR( cudaFree( dev_a0 ) );
HANDLE_ERROR( cudaFree( dev_b0 ) );
HANDLE_ERROR( cudaFree( dev_c0 ) );
HANDLE_ERROR( cudaFree( dev_a1 ) );
HANDLE_ERROR( cudaFree( dev_b1 ) );
HANDLE_ERROR( cudaFree( dev_c1 ) );
HANDLE_ERROR( cudaStreamDestroy( stream0 ) );
HANDLE_ERROR( cudaStreamDestroy( stream1 ) );
return 0;
}
首先，我使用 nvvp 进行了 CUDA 分析，发现两个内核根本没有重叠：
SO 上的一些先前帖子指出分析器可能会禁用并发内核执行，所以我做了一个简单的运行。内核循环中的总时间报告为 2.2ms，但他的分析器报告每个内核的执行时间为 1.1ms。这仍然意味着两个内核之间没有（或非常差）重叠。
我在 Tesla M2090 上使用 CUDA4.0。看来这个设备（6G）上的内核资源需求（~10s MB）应该很小，并发执行应该是实用的。不确定问题出在哪里。我应该做一些特别的事情来启用并发内核（一些 API，一些环境设置......）吗？